Hello everyone

 

I have this strange feeling that [math]\frac{(a+b)!}{a!+b!}\in\mathbb{N}[/math].

 

Can this be proven? If yes, how?

 

Never mind. Counterproof: (15+6)!/(15!+6!) [math]\notin\mathbb{N}[/math]

Edited January 28, 201411 yr by Function

15+6 =21

21!=51090942171709440000

 

5! = 120

6! = 720

5!+6!=840

 

51090942171709440000/840=60822550204416000

 

No fraction, so it's natural number.

 

ps. I see you wrote "5", instead of "15". Which was later reedited and fixed to "15".

Edited January 28, 201411 yr by Sensei

15+6 =21

21!=51090942171709440000

 

5! = 120

6! = 720

5!+6!=840

 

51090942171709440000/840=60822550204416000

 

No fraction, so it's natural number.

 

Made a mistake; the correction is not natural

You might like to play with this spreadsheet.

factor.zip

You might like to play with this spreadsheet.

 

Yes, very interesting, now. However, the number of decimals should be increased (as mentioned, the quotient in row 10 is not natural)

[latex]\frac{(1+3)!}{1!+3!}=\frac{24}7\notin\mathbb N[/latex].

Request: change of title to "(a! + b!) | (a+b)!"

Reason: wrong interpretation of the meaning a | b

Another option is to let either a or b equal 0.

Another option is to let either a or b equal 0.

 

With which reason?

If it's the main problem you're talking about...

[math]\left[a=0\oplus b=0\right]\Rightarrow \frac{(a+b)!}{a!+b!}\in\mathbb{N}[/math] isn't necessarily true:

[math]\frac{(5+0)!}{5!+0!}=\frac{120}{121}\notin\mathbb{N}[/math]

Edited January 28, 201411 yr by Function

 

With which reason?

If it's the main problem you're talking about...

[math]\left[a=0\oplus b=0\right]\Rightarrow \frac{(a+b)!}{a!+b!}\in\mathbb{N}[/math] isn't necessarily true:

[math]\frac{(5+0)!}{5!+0!}=\frac{120}{121}\notin\mathbb{N}[/math]

 

Yes, sorry. I should have been clearer in my language. For a = 0 or b = 0, (a + b)!/(a! + b!) is not a natural number. I was actually going to edit the post to reflect that, but figured my meaning would be clear.

Edited January 28, 201411 yr by John

 

Yes, sorry. I should have been clearer in my language. For a = 0 or b = 0, (a + b)!/(a! + b!) is not a natural number. I was actually going to edit the post to reflect that, but figured my meaning would be clear.

 

No problem

That's actually very logic without all the mess of this topic:

 

[math](a+b)\nmid (a,b)\Leftarrow \left((a\land b\in\mathbb{N})\land (b,a\neq 0)\right)[/math]

 

This states that [math]a+b[/math] does not divide natural integer [math]a[/math] or [math]b[/math] when both [math]a[/math] and [math]b[/math] are natural integers, and if respectively [math]b[/math] or [math]a[/math] isn't 0. Should it be 0, the statement is false. (you would get something like [math]a\mid a[/math], which is always true)

 

Something else:

 

[math]a\mid (a+b) \leftrightarrow a\mid b[/math]

 

(Don't know if everything (or anything, in worst case) I wrote is mathematically correct... I hope you know what I mean)

Edited January 28, 201411 yr by Function

I'm not sure what the first line means, but every natural number divides zero.

 

The second statement is true. For assume a divides a + b. Then a + b = na for some integer n. Thus b = na - a = (n - 1)a. Since n is an integer, n - 1 is an integer as well. Let m = n - 1. Then b = ma for some integer m, i.e. a divides b.

 

Now let a divide b. Then b = na for some integer n. Thus a + b = a + na = (1 + n)a. Since n is an integer, 1 + n is an integer as well. Let m = 1 + n. Then a + b = ma for some integer m, i.e. a divides a + b.

Edited January 28, 201411 yr by John

I'm not sure what the first line means, but every natural number divides zero.

 

The second statement is true. For assume a divides a + b. Then a + b = na for some integer n. Thus b = na - a = (n - 1)a. Since n is an integer, n - 1 is an integer as well. Let m = n - 1. Then b = ma for some integer m, i.e. a divides b.

 

Now let a divide b. Then b = na for some integer n. Then a + b = a + na = (1 + n)a. Since n is an integer, 1 + n is an integer as well. Let m = 1 + n. Then a + b = ma for some integer m, i.e. a divides a + b.

 

I have rewritten and explained the statement

Edited January 28, 201411 yr by Function

You might like to play with this spreadsheet.

 

I would use a formula like this

=FACT(B$1+$A6)/(FACT(B$1)+FACT($A6))-INT(FACT(B$1+$A6)/(FACT(B$1)+FACT($A6)))

With one variable in row one and the other in column one. My computer / libreoffice dies at around a=13 b=12.

 

You might get a really interesting pattern by looking for the length of repeating decimal and/or number of dp till repetition - my guess is diagonal stripes of 0 (top left to bottom right) and patches of chaotic simplicity into complexity in between

Yes doubtless the spreadsheet could be improved, but I'm glad to see some were interested.

 

It also demonstrates how sometimes you can test out some question by throwing together somehting along these lines, as an alternativwe to more formal analysis.

 

Yes doubtless the spreadsheet could be improved, but I'm glad to see some were interested.

 

It also demonstrates how sometimes you can test out some question by throwing together somehting along these lines, as an alternativwe to more formal analysis.

 

 

I find numerical rumination (aka number crunching) very useful when I cannot get a grip on the formal. Two factorials added together should have a nice pattern of prime factors - but I am yet to get it. If you exclude 1 and 0 then all sums of factorials are even and I think there will be patterns in their prime factors etc.

 

As (a+b)! will only have prime factors up to and including (a+b) and a!+b! can have prime factors that are significantly higher than a+b (ie 5!+2! = 122 with factors of 61 and 2) then there is no certainty of generating a natural number. I wonder if the density of natural number outcomes to the division becomes lower and lower as the factorials increase - I would imagine so.